Try to find the angle of launch that can hit the target 3 meters away. But with two different marbles with different masses.
Procedure:
1. Get materials.
2 balls of different mass
Catapult
Target
2. Place the target 3 meters away.
3. Place the different colored balls on the catapult.
4. Fire the catapult at different angles, using physics equations we have learned, to find the most effective angle to hit the target..
5. Repeat steps 3 and 4 with the second ball.
Data/calculations:
Yellow (small) ball: 9g
Red (big) ball: 14g
Distance from floor to top of catapult: .2 m
Distance from catapult to target: 3 m
Average
5°
10°
15°
20°
Red Ball
3.2
2.9
2.8
3
Yellow Ball
5.5
4.4
3.3
2.6
d = 3 meters
Vi = ?
Vf = 0
a = -9.8
t = not important
Solve for initial velocity:
Vf^2 = Vi^2 +2a*d
0 = x^2 + -19.6*3
7.66812 m/s = Vi
a = angle of degree
sin(2a) = acceleration * distance/initial velocity^2
sin(2a) = -9.8 * 3/7.67^2
at angle 14.9919
Reflection/conclusion:
We didn’t have the most accurate lab equipment but we got close with our data. Through trail, error, and a few calculations we were able to find what degrees it took to have two different massed balls hit the same target. Through the data we collected, we were able to find out that, no matter what degrees it was, the red ball always seemed to have landed very close to 3 meters. The yellow ball, on the other hand, it’s distance became smaller as the angle of launch was raised.
Title: Catapult Challenge
Lab goal/question:
Try to find the angle of launch that can hit the target 3 meters away. But with two different marbles with different masses.
Procedure:
1. Get materials.
2. Place the target 3 meters away.
3. Place the different colored balls on the catapult.
4. Fire the catapult at different angles, using physics equations we have learned, to find the most effective angle to hit the target..
5. Repeat steps 3 and 4 with the second ball.
Data/calculations:
Yellow (small) ball: 9g
Red (big) ball: 14g
Distance from floor to top of catapult: .2 m
Distance from catapult to target: 3 m
d = 3 meters
Vi = ?
Vf = 0
a = -9.8
t = not important
Solve for initial velocity:
Vf^2 = Vi^2 +2a*d
0 = x^2 + -19.6*3
7.66812 m/s = Vi
a = angle of degree
sin(2a) = acceleration * distance/initial velocity^2
sin(2a) = -9.8 * 3/7.67^2
at angle 14.9919
Reflection/conclusion:
We didn’t have the most accurate lab equipment but we got close with our data. Through trail, error, and a few calculations we were able to find what degrees it took to have two different massed balls hit the same target. Through the data we collected, we were able to find out that, no matter what degrees it was, the red ball always seemed to have landed very close to 3 meters. The yellow ball, on the other hand, it’s distance became smaller as the angle of launch was raised.
Go Physics!